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Predict the rate law from the following two-step mechanism: A2+2AB→→A2A2B(slow)(fast) Express your answer in terms of k, [A] and [B] as necessary.

User Faren
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Final answer:

The rate law for a two-step mechanism with A2 and 2AB reacting is most likely determined by the slow step and could be expressed as rate = k[A2][AB]^2. This needs experimental confirmation to ensure accuracy.

Step-by-step explanation:

The student is asking for the prediction of the rate law based on a given mechanism. In chemical kinetics, if a two-step mechanism is provided, the rate law is generally determined by the rate-determining (or slowest) step. Given a mechanism where A2 and 2AB react to form products through a slow step followed by a fast step, we will focus on the slow step to predict the rate law.

For the proposed reaction, assuming that A2 reacts with AB in a slow step to form a complex A2A2B, which then reacts rapidly in the next step, the rate law can be predicted to be dependent on the concentration of A2 and AB. However, the given mechanism is not balanced and does not provide clear stoichiometric coefficients. If we assume, based on typical mechanism conventions, that 1 molecule of A2 reacts with 2 molecules of AB, the rate law for a slow step involving a bimolecular reaction between these two species would likely be:

rate = k[A2][AB]^2

It is crucial to remember that the actual rate law and reaction order must be confirmed experimentally.

User Robert Watkins
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